What are the velocities of the balls after an elastic collision?

[lemon]

1. A solid ball A of mass 10kg traveling to the right at 2m/s collides with a solid ball B of mass 5kg traveling to the left at 5m/s. The collision is elastic. Find the velocities of the balls after the collision



2. Momentum before collision = Momentum after collision



3. (10kg x x2m/s) - (5kg x 5m/s) = 10v + 5v
-5 = 15v
v = -5/15 = -0.333m/s

Could somebody check this please?

 

[Fightfish]

No, it is not the case that the balls have identical velocities after the collision - this only happens when they coalesce together in a perfectly inelastic collision. So all you can conclude is that
$$(10)(2) + (5)(-5) = 10(v_{1}) + 5(v_{2})$$
To get further, you have to appeal to either conservation of energy or use the shortcut notion that for elastic collisions, relative speed of approach = relative speed of separation (which itself is a result obtained when invoking COE).

 

[lemon]

So if it's an elastic collision some kinetic energy has been lost - transferred to other forms - like heat and deformation.
errr...
1/2mv^{2 + -1/2mv2} = 1/2mv² + 1/2mv²

??

 

[lemon]

the second -1/2mv should be normal size

 

[Fightfish]

So if it's an elastic collision some kinetic energy has been lost - transferred to other forms - like heat and deformation.

No! That's an inelastic collision. An elastic collision is one in which no kinetic energy is lost at all. So, total initial kinetic energy = total final kinetic energy. This gives:
$$\frac{1}{2}(10)(2)^{2} + \frac{1}{2}(5)(-5)^{2} = \frac{1}{2}(10)(v_{1})^{2} + \frac{1}{2}(5)(v_{2})^{2}$$

Solve simultaneously with the COLM equation for v1 and v2.

 

[aim1732]

the second -1/2mv should be normal size

In an elastic collision there is no net change in K.E.
Why don't you just try to equate initial and final K.E and momentum using the two unknown velocities and the two known ones?

 

[lemon]

yes of course - elastic.
20+62.5 = 5v₁² + 2.5v₂²
82.5 = 5v₁² + 2.5v₂²

but I have two velocities in the same equation
how can this be solved for each one?

 

[Fightfish]

You still have $$-5 = 10v_{1} + 5v_{2}$$ from conservation of momentum

 

[aim1732]

Yeah Fightfish's right...

 

[lemon]

I'm really lost now

 

[aim1732]

I'm really lost now

Why don't you try the equating relative velocities before and after collision?
It is a two variable and two eqn. problem?

 

[Fightfish]

I'm really lost now

You have two equations:
$$82.5 = 5v_{1}^{2} + 2.5v_{2}^{2}$$
$$-5 = 10v_{1} + 5v_{2}$$
Solve them simultaneously to obtain v1 and v2.

 

[lemon]

I can't solve it simultaneously. If I double the first I loose both variables. If I half the second I loose both also. If I solve by substitution the same happens

 

[Fightfish]

I can't solve it simultaneously. If I double the first I loose both variables. If I half the second I loose both also. If I solve by substitution the same happens

Err...
$$-5 = 10v_{1} + 5v_{2}$$
$$- 2v_{1} - 1 = v_{2}$$

$$82.5 = 5v_{1}^{2} + 2.5v_{2}^{2}$$
$$33 = 2v_{1}^{2} + v_{2}^{2} = 2v_{1}^{2} + (- 2v_{1} - 1)^2 = 6v_{1}^{2} + 4v_{1} + 1$$
$$3v_{1}^{2} + 2v_{1} - 16 = 0$$

I guess that is sufficient for you to carry on?

 

[lemon]

(3v₁-8)(v₁+2)
v₁=8/3 or -2

-5=10(8/3)+5v₂
v₂=-19/3 or -6.3

-5=10(-2)+5v₂
v₂=3

when -6.3
v₁=-2.65

when 3
v₁=2

How do I know which to choose, as the direction the balls went could be any?

 

[Fightfish]

I apologise. I made a slight careless mistake above. The mistake in that post has since been corrected, and you may carry on.

 

[lemon]

I don't understand. How can I finish this problem? anyone?

 

[Fightfish]

Do the same thing as you did again in #15 except using the corrected version of the equation that I gave in #14 (edited in the post itself). You should take this opportunity as well to see how to approach such questions.

For your reference, the corrected version of the equation to solve is
$$3v_{1}^{2} + 2v_{1} - 16 = 0$$
I hope you will try to understand how that equation is arrived at.

 

[lemon]

thanks.
I understand how the equation was derived - by simultaneous equations - substitution.
I worked it myself after you showed me. It was late and I guess I was tired and rushing.
All good
Thanks for your support.